|
$\dfrac b{ka^2+b^2}\leqslant\dfrac b{2\sqrt{ka^2\cdot b^2}}=\dfrac1{2a\sqrt k}$
$\begin{aligned}
m&=\min\left\{a,\dfrac b{ka^2+b^2}\right\}\\
&\leqslant \min\left\{a,\dfrac1{2a\sqrt k}\right\}\\
&\leqslant\sqrt{a\cdot\dfrac1{2a\sqrt k}}\\
&=\dfrac1{\sqrt{2\sqrt k}}
\end{aligned}$ |
|